ĐKXĐ : \(x\ge0\) ; \(x\ne4\) và \(x\ne9\)

\(A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}+1}{x-4}\)

Câu b : \(\dfrac{1}{A}< \dfrac{1}{5}\Leftrightarrow\dfrac{x-4}{\sqrt{x}+1}< \dfrac{1}{5}\Leftrightarrow5x-20< \sqrt{x}+1\Leftrightarrow5x-\sqrt{x}-21< 0\)Mysterious Person Tới đây làm sao nữa :(((

\(\text{a) }ĐKXĐ:x\ge0;x\ne4;x\ne9\\ A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+1}{x-4}\)

\(b\text{) }\dfrac{1}{A}\le\dfrac{1}{5}\\ \Leftrightarrow A\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}-5\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5\left(x-4\right)}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5x+20}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-5x+21}{x-4}\ge0\\ \Leftrightarrow\dfrac{\left(\sqrt{x}-\dfrac{1+\sqrt{421}}{10}\right)\left(\sqrt{x}-\dfrac{1-\sqrt{421}}{10}\right)}{\sqrt{x}-2}\ge0\)

Rồi lập bảng xét dấu.

ĐKXĐ: \(x\ge1\)

\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)

Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)

Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)

\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)

\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)

\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)

\(\Rightarrow-1< m\le\dfrac{1}{3}\)

Chọn C

a: ĐKXĐ: x>=0; x<>1

b: \(P=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(5\sqrt{x}-2\right)}{\sqrt{x}+3}\)

Để P=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10\sqrt{x}+4=\sqrt{x}+3\)

\(\Leftrightarrow-11\sqrt{x}=-1\)

=>x=1/121

c: \(P-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}< 0\)

=>P<2/3

Lời giải:

ĐK: \(x\geq 0; x\neq 4;x\neq 9\)

a) Ta có:

\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)

\(P=\frac{\sqrt{x}+2+(x-9)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-3)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(P=\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{x-4}\)

b)

Ta có: \(\frac{1}{P}\leq \frac{-5}{2}\)\(\Leftrightarrow \frac{x-4}{\sqrt{x}+1}\leq \frac{-5}{2}\)

\(\Leftrightarrow 2(x-4)\leq -5(\sqrt{x}+1)\)

\(\Leftrightarrow 2x+5\sqrt{x}-3\leq 0\)

\(\Leftrightarrow (2\sqrt{x}-1)(\sqrt{x}+3)\leq 0\)

\(\Rightarrow 2\sqrt{x}-1\leq 0\) (do \(\sqrt{x}+3>0\) )

\(\rightarrow x\leq \frac{1}{4}\)

Vậy \(0\leq x\leq \frac{1}{4}\)

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

a: \(Q=\dfrac{9\sqrt{x}-6-3x+2\sqrt{x}+15x+41\sqrt{x}+28-42\sqrt{x}-34}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}\)

\(=\dfrac{12x+10\sqrt{x}-12}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}\)

\(=\dfrac{12x+18\sqrt{x}-8\sqrt{x}-12}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}\)

\(=\dfrac{6\sqrt{x}\left(2\sqrt{x}+3\right)-4\left(2\sqrt{x}+3\right)}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}\)

\(=\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\)

c: Để Q là số nguyên thì \(20\sqrt{x}+30⋮5\sqrt{x}+7\)

\(\Leftrightarrow5\sqrt{x}+7\inƯ\left(2\right)\)

hay \(x\in\varnothing\)

a) ĐKXĐ: : phải là 1 biểu thức có nghĩa. b) ko có x nên ko phải tìm

Ô xin lỗi bạn, do lúc trước mình ko thấy đề nên bấm bậy, xin lỗi nhiều

\(8,\dfrac{bc}{\sqrt{3a+bc}}=\dfrac{bc}{\sqrt{\left(a+b+c\right)a+bc}}=\dfrac{bc}{\sqrt{a^2+ab+ac+bc}}\)

\(=\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{b}{a+b}+\dfrac{c}{a+c}}{2}\)

Tương tự cho các số còn lại rồi cộng vào sẽ được

\(S\le\dfrac{3}{2}\)

Dấu "=" khi a=b=c=1

Vậy

\(7,\sqrt{\dfrac{xy}{xy+z}}=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\dfrac{xy}{xy+xz+yz+z^2}}\)

\(=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{y}{y+z}}{2}\)

Cmtt rồi cộng vào ta đc đpcm

Dấu "=" khi x = y = z = 1/3

Bài 1 dùng miền giá trị , ez tự làm